This question seems tricky and I frankly don't know how to start. I will be grateful if someone can provide a solution.

We have a triangle $ABC$ and there is a point $F$ on $BC$ such that $AF$ intersects the median $BD$ at $E$. If $AE=BC$ how do we prove that triangle $BEF$ is isosceles?

I think this has to do with ratios of sides, but I'm not getting any where. I drew a graph as accurately as I could and I am pretty confident that the objective of the problem is to somehow show that $EF$ equals $BF$, but I have no clue.
Thanks!

If you have a drawing in front of you, @Nitesh, you'll see that it is impossible, without more data, that $\,EF=BF\,$ ,as this would render the triangles $\,\Delta AED\,,\,\,\Delta CBD\,$ congruent, and then $\,BD=ED\,$ , which is absurd as the latter segment is a subsegment of the former.
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DonAntonioAug 28 '12 at 16:17

I added a diagram, but I don't see how AED and CBD have to be congruent if EF=BF?
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NiteshAug 28 '12 at 16:22

Oops, wrong! We have side-side-angle, not side-angle-side! The above doesn't render those triangles congruent, indeed...unless $\,AD=DC\,$ is the longest side in each respective triangle.
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DonAntonioAug 28 '12 at 17:13